Measure the voltage appearing across each resistor. If they are identical, and equal to the supply voltage, then the resistors are in parallel.
Because the two voltages are out of phase, that means that individually they peak at different times in the AC cycle, so in general if they are measured separately their sum will exceed the supply voltage, possibly by up to 41%.
I don't think you can do that, with the information provided.
A little more than 6 volts RMS, or 17 volts peak to peak (can get -6 and +6 voltage this way using a full wave rectifier). The voltage will depend strongly on your design - are you using circuitry that clamps the output voltage to a specific value (you should!). If you are doing this, I would find the cheapest transformer that meets the current capacity you need that has an output near the 6 volts you want.
This could be dangerous if you aren't very careful. Take the transformer out of the welding machine. Put it on a wooden bench. Apply 12 volts to the input side with a car battery. Measure the input volts with a volt meter and write down the value. Then use your volt meter to measure the output voltage. If you have no output voltage then your transformer is obviously blown. You should read an output voltage that is higher than the input voltage based on the number of winding in the coil and/or the manufacturers specifications. According to Faraday's law the only output voltage you will read is at the exact time power is applied to or taken away from the input when using dc power. Otherwise you will have to apply ac power which is much more powerful ( and more dangerous ) than the 12v dc. Your volt meter may not be designed to read high enough voltage for the output you would see with 110 v ac.
string efficiency= voltage across the string ---------------------------------------- (N*voltage across the most stressed unit) N is the number of units in the string
string efficiency= voltage across the string ---------------------------------------- (N*voltage across the most stressed unit) N is the number of units in the string
db gain is defined as power gain, not voltage gain. Please restate you question in terms of power, or provide details of input and output impedance.
Tell me what car the fuel tank is located in and I will tell you the fuel capacity.
Measure the voltage appearing across each resistor. If they are identical, and equal to the supply voltage, then the resistors are in parallel.
A dry cell consists of two different metals separated by an electrolyte, allowing for the flow of electrons. This difference in metals creates a potential difference between the terminals, resulting in an electric potential that can be measured using a voltmeter.
Solar cells produce electricity rather than store voltage. The voltage produced by a solar cell depends on factors like sunlight intensity and cell characteristics. You can measure the voltage generated by a solar cell using a multimeter or similar device.
A: Simply by adding a series resistor from the battery charger. WHAT VALUE? find the current required and use it to IR drop the voltage
Because the two voltages are out of phase, that means that individually they peak at different times in the AC cycle, so in general if they are measured separately their sum will exceed the supply voltage, possibly by up to 41%.
P(watts) = I (current) times E (voltage) hence it is known as the pie formula P=IE
Yes, a rectifier can be used to convert AC into DC.Using a single rectifier, half of a sine wave voltage can be clipped off, which leaves the other half. The voltage not steady as only half of the sine wave appears at the output during one cycle. This is half-wave rectification. A more complicated rectifier arrangement can force both halves of the AC input to be delivered in the output. One half of the sine wave will be "flipped over" so the output voltage does not change polarity.In neither case will the output voltage be "steady" and further "smoothing" circuitry will be required to make the DC output. This is the function of filtering, and many DC supplies will incorporate this feature. Lastly, we also find that DC supplies often incorporate some kind of regulation, which makes the output voltages resistant to changes when the input voltage changes and/or the load on the supply changes.
The fill factor of a photovoltaic cell is a measure of its efficiency in converting sunlight into electricity. It is the ratio of the maximum power output of the cell to the product of its open-circuit voltage and short-circuit current. A higher fill factor indicates a more efficient solar cell.